We conduct an experiment where there are only four possible outcomes: A, B, C, o

Question

We conduct an experiment where there are only four possible outcomes: A, B, C, or D. There are four possible distributions on these outcomes corresponding to theta = 0, 1, 2, or 3 respectively. These distributions are I want a test that decides between the null hypothesis theta = 0 versus the alternative theta notequalto 0 (or, in other words, the alternative that theta is either 1, 2, or 3). Consider the test that has the critical region {B}. Calculate the level of this test, and calculate the power under each of the alternatives theta = 1, 2, or 3. Find the LRT for testing H_0: theta = 0 versus H_A: theta notequalto 0 with level alpha = 0.25. Is it possible for the test with critical region {B} to be more powerful than the LRT? Explain your reasoning.

Explanation / Answer

a) The test that has the critical region {B}

Level of the test means the probability under NULL hypothesis i.e. Theta =0

so level of test=PTheta =0 (B)=0.25

Power under each alternative is the probability under Theta =1,2, or 3

P1 (B)= 0.25

P2 (B)=0.13

P3 (B)=0.1

b)

In LRT testing it takes smallest values of the ratio P0(omega)/(maxTheta PTheta(omega))

For Theta=A

P0(omega)=0.25 Probability at 0

from other probability of A for 1,2 or 3 are 0.5,0.12,0.8

max Theta is 0.8

So max Theta PTheta(omega) =0.8

P0(omega) /(max Theta PTheta(omega) )=0.25/0.8

P0(omega) /(max Theta PTheta(omega) ) =0.3125 for A

For Theta=B

P0(omega)=0.25 Probability at 0

from other probability of B for 1,2 or 3 are 0.25,0.13,0.1

max Theta is 0.25

So max Theta PTheta(omega) =0.25

P0(omega) /(max Theta PTheta(omega) )=0.25/0.25

P0(omega) /(max Theta PTheta(omega) ) =1 for B

For Theta=C

P0(omega)=0.25 Probability at 0

from other probability of C for 1,2 or 3 are 0.13,0.25,0.05

max Theta is 0.25

So max Theta PTheta(omega) =0.25

P0(omega) /(max Theta PTheta(omega) )=0.25/0.25

P0(omega) /(max Theta PTheta(omega) ) =1 for C

For Theta=D

P0(omega)=0.25 Probability at 0

from other probability of D for 1,2 or 3 are 0.12,0.5,0.05

max Theta is 0.5

So max Theta PTheta(omega) =0.5

P0(omega) /(max Theta PTheta(omega) )=0.25/0.5

P0(omega) /(max Theta PTheta(omega) ) =0.5 for D

From the above calculation the smallest ratio of P0(omega) /(max Theta PTheta(omega) ) is for A i.e. 0.3125.

So, the test with critical region {A} has smallest ratio, and with level= 0.25.

c)

Yes, it is possible, Notice when Theta = 2, the power of the test with critical region (B) from (a) is 0.13. however, the power of the LRT is only 0.12.

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